Mathematics, 16.04.2020 00:20 roro67

A survey was conducted in a large city to investigate public opinion on banning the use of trans fats in restaurant cooking. A random sample of 230 city residents with school-age children was selected, and another random sample of 341 city residents without school-age children was also selected. Of those with school-age children, 94 opposed the banning of trans fats, and of those without school-age children, 147 opposed the banning of trans fats. An appropriate hypothesis test was conducted to investigate whether there was a difference between the two groups of residents in their opposition to the banning of trans fats. Is there convincing statistical evidence of a difference between the two population proportions at the significance level of 0.05

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Answer from: jenfantroy9892

There is no enough evidence to claim that there is a difference between the two population proportions.

Step-by-step explanation:

We have to perform an hypothesis testing for a difference between two population proportions.

The null hypothesis will state that both proportions are the same, and the alternative hypothesis will state that they differ. This would be than a two-side hypothesis test.

We can write this as:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2\neq0$

The significance level for this test is 0.05.

The sample of city residents with school-age children has a sample size n1=230 and a sample proportion p1=0.41

p_1=X_1/N_1=94/230=0.41

The sample of city residents without school-age children has a sample size n2=341 and a sample proportion p2=0.51

p_2=X_2/N_2=147/341=0.43

The weighted p, needed to calculate the standard error, is the weighted average of both sample proportions:

$p=\dfrac{n_1p_1+n_2p_2}{n_1+n_2}=\dfrac{230*0.41+341*0.51}{230+341}=\dfrac{94+147}{571} =\dfrac{241}{571}=0.42$

The standard error of the difference of proportions can now be calculated as:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.42*0.58}{230}+\dfrac{0.42*0.58}{341}}\\\\\\\sigma_p=\sqrt{\dfrac{0.2436}{230}+\dfrac{0.2436}{341}}=\sqrt{0.00106+0.00071}=\sqrt{0.00177}\\\\\\\sigma_p=0.042$

The test statistic z is:

$z=\dfrac{p_1-p_2}{\sigma_p}=\dfrac{0.41-0.43}{0.042}= \dfrac{0.020}{0.042}= 0.476$

The P-value for this two side test and this value of the z-statistic is:

P-value=2*P(z0.476)=0.634

The P-value is bigger than the significance level, so the effect is not significant. The null hypothesis failed to be rejected.

There is no enough evidence to claim that there is a difference between the two population proportions.

Answer from: Quest

step-by-step explanation:

7+3=10

Answer from: Quest

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?

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A survey was conducted in a large city to investigate public opinion on banning the use of trans fat...

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